Question

The equilibrium constant for the gas phase reaction: N2O5(g) ---> 2 NO2(g) + ½ O2(g) is 95 at 25ºC. What is the value of the equilibrium constant for the following reaction at 25ºC?
O2(g) + 4 NO2(g) ---> 2 N2O5(g)


1/95
(95)^2
1/(95)^2
(95)^½

Answer 1

Answer: The value of equilibrium constant for reverse reaction is $(\frac{1}{95})^2$

Explanation:

The given chemical equation follows:

$N_2O_5(g)\rightarrow 2NO_2(g)+\frac{1}{2}O_2(g)$

The equilibrium constant for the above equation is 95.

We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:

$O_2(g)+4NO_2(g)\rightarrow 2N_2O_5(g)$

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

If the equation is multiplied by a factor of '2', the equilibrium constant of the reverse reaction will be the square of the equilibrium constant  of initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{eq}'=(\frac{1}{95})^2$

Hence, the value of equilibrium constant for reverse reaction is $(\frac{1}{95})^2$